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Is $\text{Gal} (\overline{\Bbb Q}/\Bbb Q)$ countable or uncountable? It seems like it should be countable (because the algebraic closure of $\Bbb Q$ is countable and there are countably many permutations of the irrational algebraic numbers, and a countable union of countable sets is countable. However, I've seen references that seem to imply it is uncountable. What is the answer, and why/how?

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  • $\begingroup$ It's uncountable. There are uncountably many permutations of the algebraic numbers. $\endgroup$ Apr 21 '16 at 6:04
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    $\begingroup$ @QiaochuYuan, there are uncountably many permutations of the complex numbers and yet the Galois group of $\mathbb C$ over $\mathbb R$ is finite! $\endgroup$ Apr 21 '16 at 6:10
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    $\begingroup$ @Mariano: yes, of course this isn't an argument, I just wanted to point out that the OP's claim that there are countably many such permutations is false. $\endgroup$ Apr 21 '16 at 6:14
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    $\begingroup$ There are no countably infinite Galois groups. $\endgroup$
    – Lubin
    Apr 21 '16 at 19:48
  • $\begingroup$ @Rob: related: math.stackexchange.com/questions/260223/… $\endgroup$
    – Watson
    Aug 29 '16 at 16:05
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Let $I\subseteq \Bbb N$ be any subset and let $K_I=\Bbb Q(\{\sqrt{p_i}\}_{i\in I})$ where $p_i$ is the $i^{th}$ prime. Then there are precisely $2^{\Bbb N}$ in fact the Galois group of the compositum of this extension is exactly isomorphic to

$$\prod_{i\in\Bbb N}\Bbb Z/2\Bbb Z$$

and this of course indicates there are uncountably many elements in $\text{Gal}(\overline{\Bbb Q}/\Bbb Q)$

You can also do a simple cardinality argument using inverse limits, but that technology is a bit stronger.

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    $\begingroup$ Precisely $2^{\mathbb{N}}$ what, exactly? $\endgroup$ Apr 21 '16 at 6:13
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    $\begingroup$ @Rob: elements of the Galois group of this subextension. $\endgroup$ Apr 21 '16 at 6:15
  • $\begingroup$ Excellent answer. Thanks for the easy to understand explanation! $\endgroup$ Apr 21 '16 at 6:16
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Let me elaborate on Lubin's comment: there are no countably infinite Galois groups. This follows from a general statement about topological spaces: If $X$ is compact, Hausdorff with no isolated points, then $X$ is uncountable. For the proof, see here: https://proofwiki.org/wiki/Compact_Hausdorff_Space_with_no_Isolated_Points_is_Uncountable/Lemma

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